PASASAGE:

An airline has a certain free luggage allowance & charges for excess luggage at a fixed rate per kg. Two passengers Raja & Praja have 60 kg of luggage between them, and are charged Rs.1200 & Rs.2400 for excess luggage. Had the entire luggage belonged to one of them , the excess luggage charge would have been Rs. 5400.

Q.1 What is the weight of the Praja’s luggage ?

A) 20 kg

B) 35 kg

C) 40 kg

D) 30 kg

Ans. B

Let free luggage allowed be x kg

Raja & Praja are allowed free luggage of 2xkg

Let charge for excess luggage= Rs. Y /kg

Raja & Praja pay Rs. 1200 + Rs. 2400 = 3600 excess luggage

Y(60-2x) = 3600–i

If entire luggage had belonged to one of them , excess luggage charges would have been Rs.5400

Y (60-x) = 5400—ii

From these two equations:

60y- 2xy = 3600

60y – xy =5400

Now. Solve these two equations;

Y = 120

X = 15 kg

Since Praja is charged Rs.2400 for excess luggage is 2400/120 = 20kg

Hence weight of Praja’s luggage + her excess luggage

= 15 + 20kg

=35 kg

Q.2 What is the free luggage allowance per person ?

A) 10 kg

B) 15 kg

C) 20 kg

D) 30 kg

Ans. B

Q.3 Two boys begin together to write out a booklet containing 8190 lines. The first boy starts with the first line , writing at the rate of 200 lines an hour & the second boy starts with the last line, then writes 8189th line & so on, proceeding backward at the rate of 150 lines an hour. At what line will they meet ?

A) 4910

B) 4680

C) 4280

D) 5200

Ans. B

Relative speed = 200 + 150 = 350 lines/hr

So, they will meet after =8190 / 350 = 117/5 hrs

Therefore, they will meet at 117/5*200 = 4680th line from the beginning

Q.4 If 20% of P = 30% of Q = 1/6 of R, then P:Q:R is:

A) 3 : 2: 16

B) 2 :3 : 16

C) 15 : 10 : 18

D) 10 : 15 : 18

Ans. C

PQR=1/5:3/10:1/6

=15:10:18

Q.5 A local train is going from Delhi to Saharanpur & it has 4 stoppages in between. If two passengers enter in the train during the journey , how many different sets of tickets they may had ?

A) 90

B) 120

C) 45

D) 180

Ans. C

1st station = 4tickets

2nd station = 3 tickets

3rd station = 2 tickets

4th station = 1 ticket

Total no. of tickets = 4+3+2+1 = 10

Therefore, sets of ticket they may had = 10C2 = 45

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